Problem: A few families took a trip to an amusement park together. Tickets cost $$7.00$ each for adults and $$3.50$ each for kids, and the group paid $$63.00$ in total. There were $6$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Explanation: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${7x+3.5y = 63}$ ${x = y-6}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-6}$ for $x$ in the first equation. ${7}{(y-6)}{+ 3.5y = 63}$ Simplify and solve for $y$ $ 7y-42 + 3.5y = 63 $ $ 10.5y-42 = 63 $ $ 10.5y = 105 $ $ y = \dfrac{105}{10.5} $ ${y = 10}$ Now that you know ${y = 10}$ , plug it back into ${x = y-6}$ to find $x$ ${x = }{(10)}{ - 6}$ ${x = 4}$ You can also plug ${y = 10}$ into ${7x+3.5y = 63}$ and get the same answer for $x$ ${7x + 3.5}{(10)}{= 63}$ ${x = 4}$ There were $4$ adults and $10$ kids.